Problem: Suppose that $x$ is an integer that satisfies the following congruences: \begin{align*}
3+x &\equiv 2^2 \pmod{3^3} \\
5+x &\equiv 3^2 \pmod{5^3} \\
7+x &\equiv 5^2 \pmod{7^3}
\end{align*}What is the remainder when $x$ is divided by $105$?
Solution: Since $105 = 3 \cdot 5 \cdot 7$, by the Chinese Remainder Theorem, it suffices to find the remainders when $x$ is divided by $3$, $5$, and $7$. As $3+x$ leaves a remainder of $4$ when divided by $27 = 3^3$, it follows that $3+x \equiv 4 \pmod{3}$, and thus that $x\equiv 1 \pmod{3}$. Similarly, \begin{align*}
x &\equiv 9 \equiv 4 \pmod{5} \\
x &\equiv 25 \equiv 4 \pmod{7}.
\end{align*}As $4 \equiv 1 \pmod{3}$, it follows from the Chinese Remainder Theorem that $x \equiv \boxed{4} \pmod{105}$.